The product of two consecutive integers is 38 more than 12 times their sum

Question

Solve by factoring. The product of two consecutive integers is 38 more than 12 times their sum. Find all possible integers. 

Answer 1

x(x+1)=38+12(x+[x+1])

x^2+x=38+12(2x+1)

x^2+x=38+24x+12

x^2+x=50+24x

x^2+x-24x-50=0

x^2-23x-50=0

factor

(x-25)(x+2)=0

x-25=0

x=25

x+1=26

25 and 26 are the numbers

I hit the wrong button and didn't finish the problem. x+2=0 x=-2 x+1=-2+1=-1 -2 and -1 make up the second pair of numbers

Answer 2

Let the two consecutive integers be x and x+1

so x(x+1) = 12(x + (x+1)) + 38

Distribute and expand, rewrite into a quadratic. Solve and you'll get two possible solutions.